Find $\int \dfrac{1}{\sqrt{-x^2+12x-32}}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\text{arctan}\left(\dfrac{x-6}{2}\right)+C$ (Choice B) B $\dfrac{1}{12}\text{arcsin}\left(\dfrac{x-6}{2}\right)+C$ (Choice C) C $\dfrac{1}{12}\text{arctan}\left(\dfrac{x-6}{2}\right)+C$ (Choice D) D $\text{arcsin}\left(\dfrac{x-6}{2}\right)+C$
The integrand is in the form $\dfrac{1}{\sqrt{p(x)}}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as $ k^2-(x+ h)^2$. Then, we will be able to integrate using our knowledge of the derivative of the inverse sine function: $\int \dfrac{1}{\sqrt{ k^2-x^2}}\,dx=\text{arcsin}\left(\dfrac{x}{ k}\right)+C$ [Why is this formula true?] By setting $u=x+ h$ and using $u$ -substitution, we get the following formula: $\int \dfrac{1}{\sqrt{ k^2-(x+ h)^2}}\,dx=\text{arcsin}\left(\dfrac{x+ h}{ k}\right)+C$ We start by rewriting $p(x)$ as $ k^2-(x+ h)^2$ : $\begin{aligned} -x^2+12x-32&=-32-(x^2-12x) \\\\ &=-32+36-(x^2-12x+36) \\\\ &=4-(x-6)^2 \\\\ &={2}^2-(x{-6})^2 \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{\sqrt{-x^2+12x-32}}\,dx \\\\ &=\int\dfrac{1}{\sqrt{{2}^2-(x{-6})^2}}\,dx \\\\ &=\text{arcsin}\left(\dfrac{x{-6}}{{2}}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{\sqrt{-x^2+12x-32}}\,dx=\text{arcsin}\left(\dfrac{x-6}{2}\right)+C$